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26 April, 15:03

An object is dropped from a bridge. A second object is thrown downwards 1.0 s later. They both reach the water 20 m below at the same instant. What was the initial speed of the second object

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  1. 26 April, 15:07
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    vi = 14.610

    Explanation:

    initial velocity of the first object (vi) = 0 m/s, because it was dropped

    distance (y) = - 20 m

    initial position (y0) = 0 m

    acceleration due to gravity (g) = - 9.8 m/s^2

    y = 1/2 gt^2 + vi*t + y0

    -20 = 1/2 (-9.8) t^2 + 0 + 0

    -20 = - 4.9t^2

    4.081 = t^2

    +√4.081 = t

    t = 2.020

    time of second object = 2.020 - 1 = 1.020

    Now we can plug in the new time to solve for vi of the second object.

    y = 1/2 gt^2 + vi*t + y0

    -20 = 1/2 (-9.8) 1.020^2 + vi*1.020 + 0

    -20 = - 5.098 + 1.020vi

    14.902 = 1.020vi

    vi = 14.610 m/s
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